[moving to haskell-cafe]

Hello I have little problem. I want to solve this equation

f(n-1) = f(n)+n f(6)=6

This is a simple recursive definition.. By the way, are you sure you didn't meant: f(n-1)=f(n)+(n-1)

=> f(2)=?

Not exactly sure what you're after here. You could simply do: f(2) =f(3)+3 =f(4)+4+3 =f(5)+5+4+3 =f(6)+6+5+4+3 =6+6+5+4+3 =24 anyway it's simple to see, or even prove by induction, that f(n) = (\sum_{i=n+1}^{7} i) -1 = (n+8)(7-n)/2 - 1 If I'm rigth and you're formula is suposed to be f(n-1) = f(n)+n it is even easier. J.A.